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Toolkit 42

Angle Bisector Theorem

Angle Bisector Theorem figure
BDDC=cb\frac{BD}{DC}=\frac{c}{b}

Proof

Angle Bisector Theorem figure

Draw a line through BB parallel to ADAD, intersecting the extension of ACAC at EE, as shown.

Angle Bisector Theorem construction with parallel line

Since AD∥BEAD \parallel BE,

∠AEB=∠CAD=A2,\angle AEB = \angle CAD = \frac{A}{2},
∠ABE=∠BAD=A2.\angle ABE = \angle BAD = \frac{A}{2}.

Therefore,

AE=AB=c.AE = AB = c.

By the Triangle Proportionality Theorem (Thales' Theorem),

CDDB=CAAE=bc.\frac{CD}{DB} = \frac{CA}{AE} = \frac{b}{c}.

Therefore,

BDDC=cb.□\frac{BD}{DC} = \frac{c}{b}. \quad\square